Question

A 0.160 kg hockey puck is moving on an icy, frictionless,horizontal surface. At t = 0 the puck is moving to theright at 2.40 m/s.
(a) Calculate the magnitude of the velocity of the puck after a force of 37.0 N directed to the right has been applied for 0.050 s.
a. +x direction.
b. +y direction.
c. -x direction.
d. -y direction.
(b) If instead, a force of 13.3 Ndirected to the left is applied from t = 0 to t =0.050 s, what is the final velocity of the puck?
a. +x direction.
b. +y direction.
c. -x direction.
d. -y direction.

Answer 1

Answer:

a. 13.96 m/s in the +x direction

b. -1.76 m/s that is 1.76 m/s in the -x direction

Explanation:

(a) Calculate the magnitude of the velocity of the puck after a force of 37.0 N directed to the right has been applied for 0.050 s.

Since Impulse, I = Ft = m(v₂ - v₁) where F = force applied = + 37.0 N(positive since it is applied to the right), t = duration of applied force = 0.050 s, m = mass of hockey puck = 0.160 kg, v₁ = initial velocity of hockey puck =  + 2,40 m/s (positive since it is moving to the right) and v₂ = final velocity of hockey puck

So, making v₂ subject of the formula, we have

v₂ = v₁ + Ft/m

substituting the values of the variables into the equation, we have

v₂ = v₁ + Ft/m

v₂ = 2.40 m/s + 37.0 N × 0.050 s/0.160 kg

v₂ = 2.40 m/s + 1.85 Ns/0.160 kg

v₂ = 2.40 m/s + 11.56 m/s

v₂ = 13.96 m/s

(b) If instead, a force of 13.3 N directed to the left is applied from t = 0 to t =0.050 s, what is the final velocity of the puck?

Since Impulse, I = Ft = m(v₂ - v₁) where F = force applied = - 13.3 N(positive since it is applied to the right), t = duration of applied force = 0.050 s, m = mass of hockey puck = 0.160 kg, v₁ = initial velocity of hockey puck =  + 2,40 m/s (positive since it is moving to the right) and v₂ = final velocity of hockey puck

So, making v₂ subject of the formula, we have

v₂ = v₁ + Ft/m

substituting the values of the variables into the equation, we have

v₂ = v₁ + Ft/m

v₂ = 2.40 m/s + (-13.3 N) × 0.050 s/0.160 kg

v₂ = 2.40 m/s - 0.665 Ns/0.160 kg

v₂ = 2.40 m/s - 4.156 m/s

v₂ = -1.756 m/s

v₂ ≅ -1.76 m/s

Answer 2

(a). The velocity of the puck is 13.96 m/s in the +x direction

(b). The final velocity of the puck is -1.76 m/s which is 1.76 m/s in the -x direction

What is velocity?

The velocity of the object is defined as the movement of the object in a particular direction with respect to time.

(a) Calculate the magnitude of the velocity of the puck after a force of 37.0 N directed to the right has been applied for 0.050 s.

It is given that

Impulse, I = Ft = m(v₂ - v₁)

where F = force applied = + 37.0 N(positive since it is applied to the right),

t = duration of applied force = 0.050 s,

m = mass of hockey puck = 0.160 kg,

v₁ = initial velocity of hockey puck =  + 2,40 m/s (positive since it is moving to the right)  

v₂ = final velocity of a hockey puck

So, making v₂ subject of the formula, we have

$V_2=v_1+\dfrac{Ft}{m}$

substituting the values of the variables into the equation, we have

$v_2=2.40+\dfrac{37\times 0.050}{0.160}$

$v_2=2.40+11.56$

$v_2=13.96\ \dfrac{m}{s}$

(b) If instead, a force of 13.3 N directed to the left is applied from t = 0 to t =0.050 s, what is the final velocity of the puck?

Since Impulse, I = Ft = m(v₂ - v₁)

where F = force applied = - 13.3 N(positive since it is applied to the right),

t = duration of applied force = 0.050 s,

m = mass of hockey puck = 0.160 kg,

v₁ = initial velocity of hockey puck =  + 2,40 m/s (positive since it is moving to the right)  

v₂ = final velocity of a hockey puck

So, making v₂ subject of the formula, we have

$v_2=v_1+\dfrac{Ft}{m}$

substituting the values of the variables into the equation, we have

$v_2=2.40+\dfrac{-13.3\times 0.050}{0.160}$

$v_2=2.40-4.156$

$v+2=-1.76\ \frac{m}{s}$

Thus

(a). The velocity of the puck is 13.96 m/s in the +x direction

(b). The final velocity of the puck is -1.76 m/s which is 1.76 m/s in the -x direction

To know more about velocity follow

https://brainly.com/question/4931057